4/.
a). Fe2O3 + 3CO => 2Fe + 3CO2
b)..
m.pura t 3,36 t
Fe2O3 + 3CO => 2Fe + 3CO2
160 2x56
=> m.p = 160x3,36/2x56 = 4,8 t Fe2O3 pur
p% = m.px100/m.ip
= 4,8x100/5,50 = 87,30%
5/.
24 g n moli
C + 2H2SO4 => CO2 + 2SO2 + 2H2O
12 2
=> n = 24x2/12 = 4 moli H2SO4
stim ca Cm = niu/Vs
=> Vs = niu/Cm = 4/2 = 2 L sol. H2SO4
6/.
a).
+4 -1 +2 0
MnO2 + 4HCl => MnCl2 + 2H2O + 2Cl2
Mn(4+)__+2e-__> Mn(2+) reducere (ag. ox. MnO2)
2x Cl(1-) __-1e-__> Cl(0) oxidare (ag. red. HCl)
b).
s-au obtinut 1,792 L Cl2 la rand. 80% ... dar la 100% cat s-ar fi obtinut..?
80% .................. 1,792 L CL2
100% ..................... a = 2,24 L Cl2
n moli 2,24 L
MnO2 + 4HCl => MnCl2 + 2H2O + 2Cl2
4 2x22,4
=> n = 4x2,24/2x22,4 = 0,2 moli HCl
stim ca Cm = niu/Vs
=> Vs = niu/Cm = 0,2/2 = 0,1 L sau 100 ml sol. HCl
